Integration Volumes of revolution

FM Volumes of revolution I KUS objectives BAT Find Volumes of revolution using Integration Starter: Find these integrals 𝑥 5 𝑑𝑥 = 1 6 𝑥 6 +𝐶 ( 𝑥 +4 𝑥 3 )𝑑𝑥 = 2 3 𝑥 3/2 + 𝑥 4 +𝐶 1 𝑥 2 𝑑𝑥 =− 1 𝑥 +𝐶

Notes 1 You can use Integration to find areas and volumes y y x a b dx x a b y dx In the trapezium rule we thought of the area under a curve being split into trapezia. To simplify this explanation, we will use rectangles now instead The height of each rectangle is y at its x-coordinate The width of each is dx, the change in x values So the area beneath the curve is the sum of ydx (base x height) The EXACT value is calculated by integrating y with respect to x (y dx) For the volume of revolution, each rectangle in the area would become a ‘disc’, a cylinder The radius of each cylinder would be equal to y The height of each cylinder is dx, the change in x So the volume of each cylinder would be given by πy2dx The EXACT value is calculated by integrating y2 with respect to x, then multiplying by π. (πy2 dx) 𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥

Notes 2 𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 x y a b y x This would be the solid formed 𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 Imagine we rotated the area shaded around the x-axis What would be the shape of the solid formed? http://calculusapplets.com/revolution.html

3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2 WB A1 The region R is bounded by the curve 𝑦 =2𝑥+1 and the vertical lines x=1 and x=3 Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. 3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 1 3 2𝑥+1 2 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 1 3 4 𝑥 2 +4𝑥+1 𝑑𝑥 = 158𝜋 3 See Geogebra workbookA1

WB A2 The region R is bounded by the curve 𝑦 = 1 𝑥 , the x-axis and the vertical lines x = 1 and x =2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Write your answer as a multiple of π 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 1 2 𝑥 −1 2 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 1 2 𝑥 −2 𝑑𝑥 = 𝜋 −𝑥 −1 2 1 = 𝜋 2

WB A3 The region R is bounded by the curve 𝑦 =𝑥 𝑥 , the x-axis and the vertical lines x = 0 and x =2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Write your answer as a multiple of π 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 0 2 𝑥 3/2 2 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 0 2 𝑥 3 𝑑𝑥 = 𝜋 1 4 𝑥 4 2 0 =4𝜋

WB A4 The region R is bounded by the curve 𝑦 = 3− 𝑥 3 , the x-axis and the vertical lines x = -1 and x = -3 Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Give your answer as an exact multiple of π 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 −3 −1 3− 𝑥 3 2 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 −3 −1 3− 𝑥 3 𝑑𝑥 = 𝜋 3𝑥 − 1 4 𝑥 4 −1 −3 = 𝜋 −3− 1 4 −𝜋 −9− 81 4 = 𝜋

𝑦 =𝑥 1−𝑥 , so 𝑦 2 = 𝑥 2 (1−𝑥) 2 = 𝑥 2 −2 𝑥 3 + 𝑥 4 WB A5 The region R is bounded by the curve 𝑦 =𝑥(1−𝑥), the x-axis and the vertical lines x = 0 and x =1 Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. 𝑦 =𝑥 1−𝑥 , so 𝑦 2 = 𝑥 2 (1−𝑥) 2 = 𝑥 2 −2 𝑥 3 + 𝑥 4 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 0 1 𝑥 2 −2 𝑥 3 + 𝑥 4 𝑑𝑥 = 𝜋 1 3 𝑥 3 − 1 2 𝑥 4 + 1 5 𝑥 5 1 0 = 𝜋 30

WB A6 The region R is bounded by the curve 𝑦 =3𝑥− 𝑥 2 andand the horizontal line y=2 Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. 3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 1 2 3𝑥− 𝑥 2 2 𝑑𝑥 −𝜋 1 2 2 2 𝑑𝑥 = 47𝜋 10 −4π = 7𝜋 10

NOW DO Ex 5A 𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑥 2 − 𝑟 2 2 𝑑𝑥 𝑥 2 + 𝑦 2 = 𝑟 2 𝑦= 𝑟 2 − 𝑥 2 WB A7 Find the volume of the sphere formed when a circle of radius r, centred at the origin is rotated 2π radians about the x-axis. 𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑥 2 − 𝑟 2 2 𝑑𝑥 𝑥 2 + 𝑦 2 = 𝑟 2 𝑦= 𝑟 2 − 𝑥 2 𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑟 2 −𝑥 2 𝑑𝑥 =2𝜋 𝑟 2 𝑥− 1 3 𝑥 3 𝑟 0 =2𝜋 𝑟 3 − 1 3 𝑟 3 − 0 −0 = 4 3 𝜋 𝑟 3 NOW DO Ex 5A

One thing to improve is – KUS objectives BAT Find Volumes of revolution using Integration self-assess One thing learned is – One thing to improve is –

END